Question

Random samples of size 49 are taken from a population that has 200 elements, a mean of 180, and a variance of 196. the distribution of the population is unknown. the mean and the standard error of the distribution of sample means are

Answer 1

Answer:

The mean is 180.

The standard error is 0.286.

Step-by-step explanation:

This question can be solved using the central limit theorem.

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$.

In this problem, we have that:

The mean of the distribution of the sample means is the same as the mean of the population. This means that the mean is 180.

The standard deviation of the population is the square root of the population variance. So $\sigma = \sqrt{196} = 14$.

The standard deviation of the sampling distribution of the sample mean is $s = \frac{\sigma}{\sqrt{n}} = \frac{14}{\sqrt{49}} = 2$.

The standard error of the sampling distribution of the sample mean is $SE = \frac{s}{\sqrt{n}} = \frac{2}{7} = 0.286$